點做呢一題 part a part b ar? banghead

答案下面幅圖就係, 但如果你係想靠自己去做o既, 就試下睇下呢段字先(唔知算唔算提示到你喇)
a part我估你都係factorize唔出d野
我當你用quotient rule同chain rule無D錯野, 只要化簡一下就會得出兩舊野(一舊減另一舊)
而其中一舊經過呢個化簡, 就已經等於要prove o既o個條式中前面o個一舊
至於後面o個一舊你就睇住個答案o黎砌, 需要夾硬乘多舊野上去
乘乜野自己試下諗, 諗唔到再睇答案(已經用o左藍字表示)

b part其實無咩特別, a part計到o既話只不過照樣in跟住調下位就得
(b part本來用紅字indicate o左d terms方便去睇, 但幅圖scan出o黎麻麻, 你睇到就睇, 睇唔到就算)

(字跡潦草, 請見諒...)

唔該哂你先
下面呢題都唔係好識, 條式唔知轉做 (integrate) dy

The gradient of a curve at any point (x,y) is given by dy/dx = k ( x - 5/4 ), where k is a constant.
(a) Find the value of k if the curve passes through the points (-2,6) and (0,-12).
Find also the equation of the curve.
(b) Find the area of the region in the first quadrant bounded by the curve, the y-axis and the line 4y=13x-13.
(c) If the region in (b0 is rotated about the x-axis, find the volume generated.

Part(a)我計到 k=4, equation is y=2x^2 - 5x -10,
唔知有冇錯
graph:

JR8753 :唔該哂你先
下面呢題都唔係好識, 條式唔知轉做 (integrate) dy

The gradient of a curve at any point (x,y) is given by dy/dx = k ( x - 5/4 ), where k is a constant.
(a) Find the value of k if the curve passes through the points (-2,6) and (0,-12).
Find also the equation of the curve.
(b) Find the area of the region in the first quadrant bounded by the curve, the y-axis and the line 4y=13x-13.
(c) If the region in (b0 is rotated about the x-axis, find the volume generated.

Part(a)我計到 k=4, equation is y=2x^2 - 5x -10,
唔知有冇錯
graph:

轉做In f(x) dy唔難喎
第一條都畀左dy/dx你，直接1/左佢就得
第二條咪轉做13x=4y+13 d左佢囉

Upper bound同lower bound唔駛教喇掛。